Mathematics
tomaparker1
47

Sin x / 1 - cos x + sin x / 1 + cos x = 2 csc x Prove the identity

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Chevas948

[latex]RTP: \frac{sin(x)}{1 - cos(x)} + \frac{sin(x)}{1 + cos(x)} = 2cosec(x)[/latex] Important identities to know: [latex]sin^{2}(x) + cos^{2}(x) = 1[/latex] This can be rearranged to make either functions the subject: [latex]sin^{2}(x) = 1 - cos^{2}(x)[/latex] [latex]cos^{2}(x) = 1 - sin^{2}(x)[/latex] Thus, [latex]LHS = \frac{sinx}{1 - cos(x)} + \frac{sin(x)}{1 + cos(x)}[/latex] By simplifying into one fraction: [latex]LHS = \frac{sin(x)(1 + cos(x)) + sin(x)(1 - cos(x))}{(1 - cos(x))(1 + cos(x)}[/latex] [latex]=\frac{sin(x)[1 + cos(x) + 1 - cos(x)]}{1 - cos^{2}(x)}[/latex] [latex]=\frac{2sin(x)}{sin^{2}(x)}[/latex] [latex]=\frac{2}{sin(x)}[/latex] Now, [latex]\frac{1}{sin(x)} = cosec(x)[/latex] Thus, we can say: [latex]LHS = 2cosec(x) = RHS[/latex] (as required)

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